Question: The graph of $y = f(x)$ is shown below.

[asy]
unitsize(0.5 cm);

real func(real x) {
  real y;
  if (x >= -3 && x <= 0) {y = -2 - x;}
  if (x >= 0 && x <= 2) {y = sqrt(4 - (x - 2)^2) - 2;}
  if (x >= 2 && x <= 3) {y = 2*(x - 2);}
  return(y);
}

int i, n;

for (i = -5; i <= 5; ++i) {
  draw((i,-5)--(i,5),gray(0.7));
  draw((-5,i)--(5,i),gray(0.7));
}

draw((-5,0)--(5,0),Arrows(6));
draw((0,-5)--(0,5),Arrows(6));

label("$x$", (5,0), E);
label("$y$", (0,5), N);

draw(graph(func,-3,3),red);

label("$y = f(x)$", (3,-2), UnFill);
[/asy]

Which is the graph of $y = f(x) - 1$?

[asy]
unitsize(0.5 cm);

picture[] graf;
int i, n;

real func(real x) {
  real y;
  if (x >= -3 && x <= 0) {y = -2 - x;}
  if (x >= 0 && x <= 2) {y = sqrt(4 - (x - 2)^2) - 2;}
  if (x >= 2 && x <= 3) {y = 2*(x - 2);}
  return(y);
}

real funca(real x) {
  return(func(x) + 1);
}

real funcc(real x) {
  return(func(x) - 1);
}

for (n = 1; n <= 5; ++n) {
  graf[n] = new picture;
  for (i = -5; i <= 5; ++i) {
    draw(graf[n],(i,-5)--(i,5),gray(0.7));
    draw(graf[n],(-5,i)--(5,i),gray(0.7));
  }
  draw(graf[n],(-5,0)--(5,0),Arrows(6));
  draw(graf[n],(0,-5)--(0,5),Arrows(6));

  label(graf[n],"$x$", (5,0), E);
  label(graf[n],"$y$", (0,5), N);
}

draw(graf[1],graph(funca,-3,3),red);
draw(graf[2],shift((1,-1))*graph(func,-3,3),red);
draw(graf[3],graph(funcc,-3,3),red);
draw(graf[4],shift((-1,-1))*graph(func,-3,3),red);
draw(graf[5],shift((-1,-1))*graph(func,-3,3),red);

label(graf[1], "A", (0,-6));
label(graf[2], "B", (0,-6));
label(graf[3], "C", (0,-6));
label(graf[4], "D", (0,-6));
label(graf[5], "E", (0,-6));

add(graf[1]);
add(shift((12,0))*(graf[2]));
add(shift((24,0))*(graf[3]));
add(shift((6,-12))*(graf[4]));
add(shift((18,-12))*(graf[5]));
[/asy]

Enter the letter of the graph of $y = f(x) - 1.$
Answer: The graph of $y = f(x) - 1$ is produced by taking the graph of $y = f(x)$ and shifting down by one unit.  The correct graph is $\boxed{\text{C}}.$